动态规划之数字三角形问题(POJ1163)的两种解法

题目

代码

解法1

将每一次计算过的最大值存至数组maxSum中，避免重复计算，提高效率

#include
#include
using namespace std;

#define MAX 101
int D[MAX][MAX];
int n;
int maxSum[MAX][MAX];
int MaxSum(int i,int j){
	if(maxSum[i][j]!=-1){
		return maxSum[i][j];
	}
	if(i == n){
		maxSum[i][j] = D[i][j];
	}
	else{
		int x = MaxSum(i+1,j);
		int y = MaxSum(i+1,j+1);
		maxSum[i][j] = max(x,y)+D[i][j]; 
	}
	return maxSum[i][j];
} 

int main(){
	cin >> n;
	int i,j;
	for(i = 1;i<=n;i++){
		for(j=1;j<=i;j++){
			cin >> D[i][j];
			maxSum[i][j] = -1;
		}
	}
	cout <<MaxSum(1,1)<<endl;
}

---

解法2

使用将递归转化成递推，并只用一行数组存储最大值，时间和空间都较第一种方法更优。

#include
#include
using namespace std;

#define MAX 101
int D[MAX][MAX];
int n;
int * maxSum;
int main(){
	cin >> n;
	int i,j;
	for(i = 1;i<=n;i++){
		for(j = 1;j<=i;j++){
			cin >> D[i][j];
		}
	}
	maxSum = D[n];
	for(i = n-1;i>=1;--i){
		for(j = 1;j<=n;j++){
			maxSum[j] = max(maxSum[j],maxSum[j+1]) + D[i][j];
		}
	}
	cout << maxSum[1]<<endl;
}